# Rotation: inertial frame vs. body-fixed frame

dyn
Hi
Angular momentum L is related to the moment of inertia (MOI) , I by L= Iω
In the body-fixed frame , ie. rotating with the object then ω = 0 and so the angular momentum is zero in the body-fixed frame. Is that correct ?

If i have a thin circular ring then the MOI about the centre is given by I = MR2 where M is the total mass and R is the radius. Is this the MOI in the inertial frame or the body-fixed frame ?

Thanks

Homework Helper
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I am not 100% sure (as I don't have much experience on working on different frame of references) but I think MoI is the same for all frames of reference, it is ##\omega## that changes (in magnitude and or direction) between different frames.

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Homework Helper
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Both. The body does not need to rotate in order to have a moment of inertia just like it does not need to move in order to have mass.

Delta2
A tensor can not be equal to zero in one frame and not be equal to zero in another frame.

If you consider an angular velocity ##\omega## relative an inertial frame then the phrase "relative an inertial frame" is a part of the definition of the vector ##\omega##. Once you have defined the vector ##\omega## you can expand this vector in any frame you wish. If ##\omega\ne 0## then it is so in any frame.

If you consider an angular velocity relativ noninertial frame then this angular velocity is just another vector, not the same as the previously discussed one

Dr.D
Even in the body frame, if the body is rotating, the angular velocity vector is not zero. The choice of a reference frame is essentially just choosing a set of base vectors on which to resolve the velocity vector.

Now there may be some confusion about what it means to rotate. The best understanding is to say that rotation refers to rotation with respect to an inertial frame. If there is rotation with respect to an inertial frame, then the vector exists and is nonzero, and it may be resolved on any convenient frame, including a body frame.

dyn
The body-fixed frame rotates with the body then surely with respect to that frame the body is not rotating ? Leading to ω = 0 and thus zero angular momentum in the body-fixed frame ?

Homework Helper
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Yes. Again, draw the analogy with a body moving in a straight line. In its own reference frame, its velocity is zero.

dyn
Dr.D
Angular momentum and velocity are physical vector quantities that exist without reference to a coordinate system. dyn is relying too much on a mathematical expression rather than looking at the physical reality.

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Angular momentum and velocity are physical vector quantities that exist without reference to a coordinate system.
I am not sure I understand what you mean by "exist". A velocity vector has magnitude and direction. A coordinate system is necessary to specify the direction. If you take that reference away, the vector loses one of its two principal properties and becomes a scalar. Doesn't that mean that it loses its "existence" as a vector?

On edit: The exchange between @Dr.D and me has gone off-thread to private messaging in order to keep the main discussion focused.

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Delta2
I think that two participants of this discussion just want to stress that a vector is a geometric object independent on coordinate frames.

The body-fixed frame rotates with the body then surely with respect to that frame the body is not rotating ? Leading to ω = 0 and thus zero angular momentum in the body-fixed frame ?
angular momentum relative the body-fixed frame is zero
Yes, and this observation is physically useless

dyn and Delta2
dyn
1 - In general , it is stated that in the lab frame the MOI matrix is time-dependent but for symmetric objects there seems to be no time-dependence. Does the time-dependence only enter in when symmetry is absent ?

2 - In Euler's equations such as

τ1 = I1##\dot{ω}##1 + (I3-I22ω3

the David Morin book states that the components above are measured with respect to the instantaneous principal axes which is the body-frame. How does an observer in an inertial (lab) frame determine the angular frequencies at which he would see the rigid body rotate ?

$$\Omega\boldsymbol x=\boldsymbol\omega\times\boldsymbol x.$$
$$\dot X=-\Omega X,\quad X(0)=E.$$